Precalculus: Mathematics for Calculus, 7th Edition

by
Stewart, James; Redlin, Lothar; Watson, Saleem

Published by
Brooks Cole

ISBN 10:
1305071751

ISBN 13:
978-1-30507-175-9

Chapter 1 - Review - Exercises: 73

Answer

$a)$ $\dfrac{6}{5}+\dfrac{8}{5}i$
$b)$ $2+0i$

Work Step by Step

$a)$ $\dfrac{4+2i}{2-i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{4+2i}{2-i}=\dfrac{4+2i}{2-i}\cdot\dfrac{2+i}{2+i}=\dfrac{(4+2i)(2+i)}{2^{2}-i^{2}}=...$
$...=\dfrac{8+4i+4i+2i^{2}}{4-i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{8+4i+4i+2(-1)}{4-(-1)}=\dfrac{8+8i-2}{4+1}=\dfrac{6+8i}{5}=...$
$...=\dfrac{6}{5}+\dfrac{8}{5}i$
$b)$ $(1-\sqrt{-1})(1+\sqrt{-1})$
Evaluate the product:
$(1-\sqrt{-1})(1+\sqrt{-1})=1^{2}-(\sqrt{-1})^{2}=1-(-1)=...$
$...=1+1=2+0i$